TECHGURU CLASSES for SSC-JE/RAILWAYS/ORDINANCE
CHAPTER - 5 : CENTROIDS
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CHAPTER - 5 : CENTROIDS
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CHAPTER - 5 : CENTROIDS
5.1
INTRODUCTION
5.1.1
Centre of Gravity
Weight of an object is a body force i.e. force of gravitational attraction of the earth
acts on each particle and is distributed over entire body. Many times, Requirement
is to find a single point inside the body at which the resultant weight (i.e. resultant
of the forces of gravitational attraction acting on each and every particle) may be
assumed to act. This point is called centre of gravity. "Principle of moments" is used
to locate the centre of gravity as follows.
Consider a body of arbitrary shape and size oriented in space as shown in figure. A
particle at position (x, y, z) is selected, weight of this small particle is 'dw'. Let
position of centre of gravity i.e. the point where the resultant weight acts, is
(x, y, z) [see figure] Applying "Principle of moments" about y axis
Sum of the moments of  Moment of the resultant 

 

weights of infinitesi mal   weight about y axis

particles about y axis.  



 
 x dw  x  d w
Where integration shows the effects of
sum.
z
y

d w
C.G.
xd w
Thus
x 
.... 5.1.1(i)
dw
x
x


y d w
...5.1.1(ii)
d w
Similarly, we can get z. Coordinate of
C. G. by
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Z
Again, Applying principle of moments, about x-axis.
y. dw  y . dw  y 
W
y
y
Figure: 5.1.1(a)
x
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z 
z d w
d w
.....5.1.1(iii)
It is to be noted that C. G. is a unique point inside the body and it does not depend
upon the orientation of the body.
5.1.2
Centre of mass
Refer figure 5.1.1(a), if the body is kept in a uniform field then weight of small
element can be written as,
dw = g. dm.
Where 'dm' is the mass of the inifinitesimal particle. Substituting dw = g dm into equations 5.1.1 (i), (ii) and (iii), we get,
x 
Similary; y 
and
z 
 x g dm   x dm ,
 g dm  dm
{g, being constant, cancels out}
..... 5.1.2(i)
 y g dm   y dm
 g dm  dm
.....5.1.2(ii)
 z. g. dm   z. dm
 g. dm  dm
.....5.1.2(iii)
As, acceleration due to gravity i.e. 'g' does not appear in the expressions of x , y
and z . Equations 5.1.2(a), (b) and (c) describe position of a point which is solely a
function of distribution of mass. The point thus obtained is called "centre of mass".
Thus, centre of mass can be defined as a point inside the body where total mass of
the body can be assumed concentrated.
5.1.3
Centroids
If the body considered for instance has a uniform mass density  then mass of the
element can be expressed as,
dm =  dV
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CHAPTER - 5 : CENTROIDS
Where 'dv' is the volume of differential element of the body in figure 5.1.3(a).
Substituting dm =  dV into equations 5.1.2(i), (ii) and (iii); we get,
x  dV  x dV

 being constant, cancels out}
x


dV
dV


Similarly y 
.....5.1.3(i)
 y dV and z   z dV
dV
dV
.....5.1.3(ii) and (iii)
As, mass density  of the body does not appear in the expressions of x , y
and z , equations 5.1.3 (i), (ii) and (iii) describe the position of a point which is
completely dependent upon the geometrical shape and size of the body. The point
thus obtained is independent of the gravity and mass distribution and is called
centroid of the body. Depending upon the geometry involved, following three cases
are discussed to determine the position of centroid of a body.
(i)
Centroid of a volume : A differential element of volume 'dV' is chosen. x, y, z
are the coordinates which describe the position of the element as shown in figure:
5.1.3 (a). Let x , y and z are the coordinates describing the position of centroid.
Use following expressions to find x , y and z .
z
 x dV
x
dV
y
dv
y

dV
y dV
C.G.
Z
Z
x
x
and z 
 z dV
dV
y
Figure 5.1.3(a)
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z
(ii) Centroid of an area : If the body is in a shape of
lamina (curved or plane) of uniform thickness 't'
throughout, then the body may be considered as an
area.
y
dA
t
Consider a curved lamina shown in figure: 5.1.3(b)
Volume 'dV' of infinitesimal element can be written as.
dV = t dA,
z
z
x
O
Putting dv = t dA into equations 5.1.3(i), (ii) and (iii),
x
y
y
We get
x 
C.
 x (t dA)
 t dA

 x dA
 dA
x
.....5.1.3(iv)
Figure 5.1.3(b)
{ t, being constant throughout, cancels out }
Similarly; y 
 y dA
 dA
....5.1.3(v)
z 
 z dA
 dA
....5.1.3(vi)
and
In case, the body is in shape of a plane lamina as shown
Figure 5.1.3(c)
in figure: 5.1.3(c), only x and y need to be determined.
(iii) Centroid of a line : Sometimes, the body may be in shape of a line (curved or
straight), such as a long slender rod or wire. The cross sectional area 'A' of such a
body remains uniform throughout its length.
Volume 'dV' of differential element can be written as
dV = A 'dl'
x
Where 'dl' is the length of the element as shown in
figure 5.1.3(d). Substituting dV = A dl into equations
5.1.3 (i), (ii) and (iii).
x 
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 x Adl   x dl
 A dl  dl
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..... 5.1.3(vii)
y
Figure 5.1.3(d)
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CHAPTER - 5 : CENTROIDS
Similarly;
y 
 y dl
 dl
..... 5.1.3(viii)
Thus, in any case, we have to choose a differential element and then using suitable
expressions, we can determine centroidal coordinates.
5.2
CENTROIDS OF LINEAR CURVES
Centroid of a straight line lies at its mid point. Position of centroid of a circular arc
is determined as follows.
5.2.1
Centroid of a circular arc
A circular arc is shown in figure: 5.2.3(a). x–y
axis are chosen in such a way that arc is
symmetric about x-axis. Centroid of the arc
lies on the x–axis so that y = 0. x–coordinate of
the centroid i.e. x is calculated as follows.
r = radius of the arc, 2  = Angle subtended by
the arc at the centre. Choose on element of
length 'dl' at angular position ' '. Angle
subtended at centre by this small element is
'd  ' as shown
A
(x,y)
d
O
B
dl
x
r
dl = r. d
Coordinates of the element are (x, y)
Figure 5.2.3(a)
Using expression 5.1.3(vii) and (viii)
x
 x dl   r cos dl
dl
dl
x


 In triangle OAB, r  cos  x  r cos

 r cos. rd
x
–
dl  r. d

 rd
–
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
r

 cos  d
–


 d
r  sin – 
– 

x
r 2sin r sin

2

.....5.2.1(i)
–
Following points to be remembered :
(i)
If circular arc is symmetrically placed about
x–axis then
x=
(ii)
r sin
, y =0

r
.....5.2.1(ii)
c
(x,y)
2
If circular arc is symmetrically placed about
y–axis then
r sin
y =
, x =0

(iii)
y
Figure 5.2.1(b)
.....5.2.1(iii)
y
If 2 =  then arc becomes semicircular as
shown in figure 5.2.1(d). For a semicircular arc
symmetrically placed about x–axis. [see figure:
5.2.1(d)]
x
x
c
r sin /2 2r

and y = 0 .....5.2.1(iv)
/2

(x, y)
r
2
If, otherwise, it is symmetrically placed about y–
axis as shown in figure: 5.2.1(e), then
x
Figure 5.2.1(c)
2r
x = 0 and y 

.....5.2.1(v)
(iv) If 2  /2, then arc becomes quarter-circular as
shown in figure: 5.2.1(f). For a quarter circular
arc symmetrically placed about x and y axes,
position of the centroid can be determined by
substituting    /4 and following similar
arguments
For a quarter-circular-arc lying completely in the
first quadrant as shown in figure: 5.2.1(h) the
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c
Figure 5.2.1(d)
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y
y
r
c
c
c
x
/2
r
/2
Figure 5.2.1(e)
Figure 5.2.1(f)
x
position of the centroid is given as follows
xy 
5.2.2
2r *
.

Figure 5.2.1(g)
y
.....5.2.1(vi)
Centroid of Composite curve or wire
c
2r/
A curve consisting of two or more standard curves is called a
composite curve. One typical composite curve is shown in
figure 5.2.2(a).
Shown composite curve consists of 1.
2.
0
x
1m
C
a quarter circular arc having radius = (1m)
and centre at B (4, 0)
a horizontal straight line DE (1m).
E
1m
1m
0
4.
x
y
D
a vertical straight line CD (=1m) and
A
Figure 5.2.1(h)
a straight line OA (=3m)
3.
y
/2
3m
A
B
x
Figure 5.2.2(a)
To determine the position of centroid of a composite curve, principle of moments is
used. Following steps are followed.
Step (i): Locate the centroid of individual curves. Also note their lengths. [As, in
our example, centroid of straight line being at its midpoint and that of quarter
circular arc at x  y 
2r
with respect to its extreme radii BA and BC, centroids of


4r 2
 2r 
  2 (x ) 2   (As x  y )
* In  OAC  (x )  (y )  

 
2
2r
r
2
 x2    x 


2
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all the four parts are located
x1 = 1.5, y1 = 0, L1 = 3m
x2 = 4 –
2
× 1 = 3.55, y2 =

x3 = 4, y3 = 1 +
x4 = (3 + 1) +
2 1
2
= 0.45, L2 =
× 1 = 1.5



, L = 1m
 3

= 4.5, y4 = 1 + 1 = 2m, L4 = 1m.]

Step (ii): Apply principle of moments. Let position of the centroid is x , y and total
length of the curve is L (= L1 + L2 + L3 .............).
[Sum of moments of individual lengths] = [Moment of total length]
L1 x1 + L2 x2 + L3 x3............. = [L1 + L2 + L3 .......] x
x 
L1 x 1  L 2 x 2  L 3 x 3  ..........
=
L 1  L 2  L 3 .............
Similarly
y 
L x
L
i
i
i
L1 y 1  L 2 y 2  L 3 y 3  ..........
=
L 1  L 2  L 3 .............
L y
L
i
i
i
{For our example, principle of moments about y-axis gives
x 
4.5  5.57  4  4.5
  1.5  1.57  3.59  1  4  1  4.5
=
= 2.83 m.
.
3  1.57  1  1
and principle of moments about x-axis gives
y 
3  10  1.57  0.45  1  1.5  1  2
3  1.57  1  1

4.21
= 0.64 m}
6.57
We will, further, avoid this long procedure Instead, a table will be formed as shown
below.
S.No
Length (Li)
1
2
3
4
3
1.57
1
1
 Li = 6.57
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x–coordinate y–coordinate
of centroid (xi) of centroid (yi)
1.5
0
3.55
0.45
4
1.5
4.5
2
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L i xi
L i yi
4.5
5.57
4
4.5
 Lixi = 18.57
0
0.71
1.5
2
 Liyi = 4.21
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CHAPTER - 5 : CENTROIDS
 Li yi
and use the formulae x   L i x i and y 
to determine the position of
 Li
 Li
centroid of the composite curve.
It follows that x 
y 
and
5.3
18.57
 Li xi
=
= 2.83 m
6.57
 Li
4.21
 Li yi
=
= 0.64 m
6.57
 Li
CENTROIDS OF PLANE SECTIONS
y
5.3.1
Quadrilateral area
C(x3, y3 )
A general quadrilateral with four vertices A (x1, y1), B
(x2, y2), C (x3, y3) and D (x4, y5) is shown in figure:
5.3.1(a). coordinates of the centroid are given as,
x 
and y 
x1  x 2  x 3  x 4
4
D(x4, y4 )
G
A (x1, y1)
.....5.3.1(i)
x
Figure 5.3.1(a)
y1  y 2  y 3  y 4
4
.....5.3.1(ii)
y
(ii) Rectangular area : Rectangular area, as shown in figure, is symmetric about both x and y axes if axes are chosen
passing through mid points. It follows that,
xy0
c
x
.....5.3.1(iii)
Figure 5.3.1(b)
Centroid lies at origin.
However, if axes are chosen in such a way that the
whole rectangular area lies in first quadrant then
x
B(x2, y2)
b
d
and y 
2
2
y
.....5.3.1(iv)
c
d/2
b/2
x
Figure 5.3.1(c)
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5.3.2
Triangular area
A triangular plane area is shown in figure
y
5.3.2(a). Coordinates of the vertices of the triangle are A(x1, y1), B (x2, y2) and C(x3, y3). In gen-
B(x2, y2)
eral, the position of the centroid ( x , y ) is given as
follows.
x
y 
.G.0.
x1  x 2  x 3
3
C(x3, y3)
A
(x1, y1)
.....5.3.2(i)
x
Figure 5.3.2(a)
y1  y 2  y 3
3
.....5.3.2(ii)
Now, as follows, It is proved that, centroid of the
triangular area lies at one third of its height from
y
B
its base. Choose x-axis coinciding with the base of
the triangle
dy
bi
h = height of the triangle, b = base of the triangle.
Choose a rectangular strip (thickness = dy) at a
height of y from base (see figure). Let b1 is the width
A
b
dA =   (h – y) dy
 h

 y dA
 dA
y
b
C
x
Figure 5.3.2(b)
of the strip.
Using expression 5.1.3(iv) y 
h
Where dA = elemental area of strip = b1 dy
using concept of simlar triangle. ABC ~ BDE.
Hence b1  h
b h–y

This value of dA, when substituted into equation 5.1.3(e)
h
y
b
 y h (h – y) dy
0
h
b
 h (h – y) dy
0
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b and h, being constants, cancel out of the integrations.
h
h

y
 hy 2 y 3 
–






h

y2 
(h – y) dy
hy
–


 0

y (h – y) dy
0
h

0
h3 h3
–
3
  h   h
y 
3
h2
h2 
h2 –

y 
5.3.3
.....5.3.2(i)
h
from base.
3
Circular areas
Centroid of a circular area lies at the centre.
Centroid of a circular segment is located as follows. Consider a circular segment of radius 'r'.
The angle subtended by the arc at the centre is
2  . Choose the x–axis in such a way that seg-
B
ment is symmetric about it. Therefore y = 0.
r
3×
D
dl=rd
C
d
A differential element OCD is chosen as shown
in figure. In limit, as d  is very small, differen-
G
O
E
r
tial element OCD can be treated as a triangle.
A
Centroid of the triangle OCD is at a distance
Figure 5.3.3(a)
2r
along OG. x coordinate of centroid of differ3
ential element is given as
x = OE = OG cos  =
2r
cos 
3
Now, using expression 5.1.3 (iv)
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x 

 dA
x dA
2r
x  3



–
 2r cos    1  r  r d

 

 3
 2
 1
 r  r d
– 2


 cos d
 d
–

2r
sin–  2r 2 sin 


3

 

3  2 
– 
–
x
2r sin
3
.....5.3.3 (i)
y
Following cases are worth considering :
Case (i): Circular segment symmetrically placed
about x–axis
r
2r sin
x
,y0
3
2
x
c
.....5.3.3 (ii)
Case (ii): Circular segment symmetrically placed
Figure 5.3.3(b)
about y–axis.
y
Centroid lies on y–axis so that
x  0, y 
2r sin
3
.....5.3.3 (iii)
c
Case (iii): If 2  =  then circular segment becomes
semicircular area as shown in figure: 5.3.3(d) and
5.3.3 (e).
2
x
Figure 5.3.3(c)
4r/3
c
c
4r/3
Figure 5.3.3(d)
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CHAPTER - 5 : CENTROIDS
In case, semicircular area is symmetrically placed about x–axis as shown in figure
5.3.3(d). As       /2
r sin /2
4r
and y = 0

3/2

x
.....5.3.3(iv)
Otherwise, when it is symmetrically placed about y-axis as shown in figure: 5.3.(e)
then x = 0 and
4r

y =
.....5.3.3(v)
Case (iv): If 2  =  /     /4 then area becomes a quarter
y
circle as shown in figure 5.3.3(f). choose x and y axes in such
a way that the quarter circle lies completely in the first quadrant.
c
OC =
r sin  /  r 1

 / 
3 2
D
O
r
x
Figure 5.3.3(f)
OC =
 r
3
x = OD = y = CD
In triangle OCD, OD2 + CD2 = OC2
  r

2 OD = OC = 

3



2

x  OD 

x

2
1  r
r

3

3
2
r
3
Thus, for a quarter circular area lying completely in first quadrant
x
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CHAPTER - 5 : CENTROIDS
Example 5.1: Determine the x and y coordinates of the
trapezoidal area shown in figure by first principle.
y
C
Solution: First principle involves integration. Therefore we
must know equation of the line BC. [B = (l, b), C = (0, a)]
B
a
b
a – b
 (x – l)
y –b = 
 0–l 
O
l
x
A
Figure : Ex-5.1(a)
a – b
 y=b– 
 (x – l)
 l 
Choose a rectangular strip (see figure) as a differential element. Width of the strip is
'dx'.
 a – b

 (x – l) dx.
Area dA of the strip is given as dA = y. dx = b – 
l

 

Centroidal coordinates of the rectangular strip are –
y
y
dx
xc = x +
, yc = .

2
Use expressions x 
 x dA
 dA
c
and y 
 y dA
 dA
c
y
O
x
x and y .
dx
x
Figure : Ex-5.1(a)


to determine
dx 
  x  2 dA
x 
 dA
Neglecting higher order differential terms,
l
x

 dA
x dA

0
l

a– b
 (x – l) dx
l 



a–b
 (x – l) dx
l 

 b – 
0
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l
x
 bx 2  a – b  x 3
x2 
–
 (a – b) 


2 0
 l  3
 2
l
2


a– b x
bx
–
 (a – b) x 



 l  3

0
al 2

x 6
al

2
l
=
 ax 2  a – b  x 3 
–
 

 l  3 0
 2
l
2

a – b x 
ax
–




 l  2 0

al 2  a – b  l 2
–

2  l  3
=
l
al – (a – b)
2
bl 2
3   a  2b  l


bl
 ab  3
2



 y dA
and y  
 dA
=
l

 a – b

b –  l  (x – l) dx

 


o
l
Ans.

a – b
 (x – l) dx
l 



o
l
=
 b – 
o
l



 a – b 
a –  l  x  dx
 
 

a – b 
 x dx
l  
 a – 
o
l

   a – b   
 a – 
 x 
    l   
0
=
a – b
– 

 l 
l
2

a – b x 
 
ax – 
 l   

–l
(a 2  ab  b 2 )
 al bl 
3
3
[b
–
a
]



= 6 (a – b)
=
3 (a – b)
2 2
Ans.
Example 5.2 : Locate the centroid of the area of the
parabolic segment shown in figure: Ex-5.2(a).
y
Solution: First of all, choose a suitable differential element.
Here, a vertical rectangular strip is chosen (width = dx).
A rea of the strip is dA = y dx = kx2 dx
2
kx
y=
Now, centroid of the differential strip is located at
0
y
dx
xc = x +
and yc =


Use the expressions x 
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 dA
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Figure : Ex-5.2(a)
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 y dA
 dA
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y
determine x and y ,
x 


dx 

x c dA
  x  2 dA
 dA

dA
2
kx
y=
y
Neglecting higher order differential terms.
x
x
a

x 
 dA
a
 x (kx
x dA
2
dx)
0


a
 (kx
2
dx)
0
dx
x
3
dx
x
2
dx
Figure : Ex-5.2(b)
0
a
0
a
 x4 
 
 4 
x
a 4 /4
3a

4
a 3 /3

a
 x3 
 
 3 
Ans.
a
y 

y c dA



y
dA
1
2

2
dA

dA
 kx
2
(kx 2 dx)
0
a
 kx
2
dx
0
a
k

2
x
2
dx
0
a
x
2
 
 
a

dx
k x 5 /5 
k a 5 /5

2 x 3 /3 a
2 a 3 /3
0
3 ka 2
10
Value of k can be determined as follows :
y 
y = kx2  k =
y
x2
When x = a, y = h  k =
put k =
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a2
h
into equation of y , we get,
a2
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10
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TECHGURU CLASSES for SSC-JE/RAILWAYS/ORDINANCE
CHAPTER - 5 : CENTROIDS
Try Yourself
Q 5.1
Locate the centroid of the area bounded by two straight lines as shown in figure
using first principle.
Q 5.2
Locate the centroid of the area bounded by an ellipse
y2
x2

 1 in the first
a2
b2
quadrant as shown in figure .
y
y
3
2
x2 y

1
a2 a 2
2
1
1
2
3
x
x
Figure : Q.5.1
5.3.5
Figure : Q.5.2
Centroid of Composite area
An area made of two or more standard areas is called a
composite area. A composite section (or area) is shown y
in figure: 5.3.5(a).
1m
1m
To determine the position of centroid of composite
area, principle of moments is used, following steps are C
used.
Step (i): Identify the standard areas {As for our example, standard areas of which composite area is made
are –
1.
A rectangular area (OABC)
2.
A triangular area (CDE)
3.
A semicircular area (AB)}
3m
B
D
0
5m
1m
A
x
Figure : 5.3.5(a)
Step (ii) : Locate the centroids of each standard area with respect to chosen
coordinate system. Also calculate their areas
{For our example, coordinates of centroid of rectangle OABC are given as x1 
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= 1 m and A1 = 10 m2, that of triangle CDE – x 2 
= 1,
2
2
3
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CHAPTER - 5 : CENTROIDS
225 9
1
  3 m, A2 =
× 2 × 3 = 3 m2
3
3
2
and centroid of semicircular area
y2 
x3  5 
1
= 5.42 m
3
 (1) 2
= 1.57 m2
2
Step (iii): Apply principle of moments y 3 = 1 m and A3 =
Sum of moments of the 


areas about any axis


Moment of the sum of 


areas about same axis 
This principle, about y–axis, gives
A1 x1 + A2 x3 + A3 x3 ..... = x (A1 + A2 + A3 .......)
x
A 1 x 1  A 2 x 2  A 3 x 3  .......
A 1  A 2  A 3  ........
 x 
 Ai xi
 Ai
and about x–axis, it gives
A1 y1 + A2 y2 + A3 y3 = y (A1 + A2 + A3 .......)
y
A 1 y 1  A 2 y 2  A 3 y 3  .......
A 1  A 2  A 3  ........
{for our example,
and
y 
x 
 y 
 Ai yi
 Ai
 Ai xi
25  3  8.51

 2.5 m
 Ai
.
 Ai yi
10  9  1.57

 1.41 m }
 Ai
.
Instead of adopting the procedure explained above, a table having all the required
details may be formed as shown below :
S.No.
Areas (m2) Ai
xi
yi
Ai xi
Ai yi
1.
10
2.5
1
25
10
2.
3
1
3
3
9
3.
1.57
5.42
1
8.51
1.57
 Ai xi = 36.51
 Ai yi = 20.57
 Ai = 14.57
Now
x 
 Ai xi
.

 2.5 m
 Ai
.
and
y 
 Ai yi
.

 1.41 m
 Ai
.
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TECHGURU CLASSES for SSC-JE/RAILWAYS/ORDINANCE
CHAPTER - 5 : CENTROIDS
Following table summarizes the location of centroids of plane figures :
S.No.
Object
Illustration
1.
Straight Line
y
x
l
2.
Circular (arcs)
(i)
x
y
l
2
0
r sin

0
0
r sin 

r

0
0
r

–
–
b
2
d
2
y
r
2
x
G
(ii)
G
x
(iii)
G
(iv)
G
x
(v)
y
G
x
3.
Rectangular area
d
G
b
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4.
Triangular area
B(x2 , y2 )
A
(x1 , y1 )
5.
Circular area
x1  x 2  x 3
3
y1  y 2  y 3
3
r sin
3
0
0
r sin
3
r
3
0
0
r
3
r
3
r
3
C(x3 , y3 )
(i)
y
x
(ii)
x
(iii)
x
x
s
(iv)
x
(v)
y
x
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CHAPTER - 5 : CENTROIDS
Example 5.3: Locate the centroid of the shaded area
shown in figure: Ex-5.3(a).
y
Solution: The given area is compsed of a
rectangular area (12×10 cm2), and a triangular area
Dl
12cm
lC
4cm

2
  6  10 cm . Further, one small rectangular


hole (2 ×
4 cm2) and a semicircular hole
G
I
H
J
4cm
O
E
3cm F
2cm
2cm
A
B
2cm 5cm
l
2
 3
2


  cm  are made. While determining the posi

Figure : Ex-5.3(a)
tion of centroid of the composite section, area of small rectangular hole and that of
semicircular hole are considered negative because there areas are removed from the
total area in order to get shaded composite section.
Form a table as shown below :
S.No.
Areas (cm 2)
x i (cm)
y i (cm)
A i x i (cm 3)
A i y i (cm 3)
1.
Rectangle OACD
12 × 10 = 120 cm 2
12/2 = 16
10/2 = 5
720
600
2.
Triangle ABC
1
× 6 × 10 = 30
2
12  12  18
= 14
3
0  0  10
= 3.33

420
100
3.
Rectangle GHIJ
–2 × 4 = –8
–96
–32
4.
Semicircle EF
–   32
= –14.13

–84.78
–17.94
11 +

= 12

3 + 6/2 = 6
2+

=4

4 ×3 / 3  = 1.27
 A i  127.87 cm 2
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 Ai xi  9
 Ai yi  6
59.22 cm
50.06cm 2
2
Now
x 
 Ai xi
.

= 7.5 cm
 Ai
.
Ans.
and
y 
 Ai yi
.

= 5.08 cm
 Ai
.
Ans.
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CHAPTER - 5 : CENTROIDS
x
 x dv , y   y dv , z   z dv
 dv
 dv
 dv
5.2 Centroid of curves (wires).

Centroid of a curved wire or rod whose cross section is very small and
uniform over its length can be determined by following expressions.
x

 x dl , y   y dl , z   z dl
 dl
 dl
 dl
Centroid of a circular arc which subtends 2  at centre is given as x = r
sin 
along a line which symmetrically divides the arc into two parts. x is

measured from centre and r is the radius of arc.

Centroid of a composite curve is determined using principle of moments.
x
 Li yi
 Li xi
 Li zi
, y 
, z 
 Li
 Li
 Li
5.3 Centroid of Sections (AREAS)

A body, whose thickness is very small and uniform over entire surface, is
treated as section. Centroid of such a plane section is determined by
x 
 x dA ,
 dA
y
 y dA , z   z dA
 dA
 dA
A section or surface may be curved or may completely lie in a single plane.
z will be zero, in case a plane section completely lie in xy plane.

Centroid of a quadrilateral area is determined by x 
x1  x 2  x 3  x 4
and

y1  y 2  y 3  y 4
where (x1, y1), (x2, y2), (x3, y3) and (x4, x4) are coor
dinates of vertices with respect to a given set of axes.
y 
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TECHGURU CLASSES for SSC-JE/RAILWAYS/ORDINANCE
CHAPTER - 5 : CENTROIDS

Centroid of a triangular area is determined by x 
x1  x2  x3
4
and
y1  y 2  y 3
were (x1, y1), (x2, y2) and (x3, y3) are coordinates of
4
the vertices of triangle with respect to a given set of axes.
y 
2r sin
from centre along a
3
line which symmetrically divides the segement into two parts.

Centroid of a circular segment is given by x 

Centroid of a composite area is determined by using principle of moments.
x 
 Ai xi
,
 Ai
y 
 Ai y i
 A i zi
, and z 
 Ai
 Ai
z = 0, in case of a plane section lying completely in xy plane.
5.4
Centroids of Homogeneous Solids (Volumes)

Homogeneous solid means a body whose density is uniform over entire
mass. Centroid of such a body is determined by using following
expressions
x 

 x dV ,
 dV
y 
 y dV ,
 dV
and z 
 z dV
 dV
Centroid of a right circular cylinder lies at the center of mid section, while
that of a semicircular cylinder lies at a perpendicular distance of
4r
from
3
the base of the mid section.

Centroid of a sphere lies at its centre while that of a hemisphere lies at a
distance of
3r
from centre along a line about which the hemisphere is

symmetric.
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TECHGURU CLASSES for SSC-JE / RAILWAYS / ORDINANCE
CHAPTER - 5 : CENTROIDS

Centroid of cone lies at
h
from the base along the cone axis; h is the height

of the cone. Centroid of a half cone volume lies at

h
from the base of cone.

Centroid of a composite volume can be determined by using principle of
moments as follows.
x 
 Vi x i
,
 Vi
y 
 Vi y i
 Vi z i
, and z 
 Vi
 Vi
EXERCISES
(I)
Objective type questions
5.1 The altitude of a right circular cone is h. The height of centre of gravity of
the cone from the base will be:
A
(a)
h
2
G
(b)
h
4
(c)
h
3
(d)
2
h
3
h
y
+
r
Figure: 5.1
5.2 The centroid of a triangle with altitude h with respect to its base is :
(a)
h
2
(b)
2h
5
(c)
h
3
(d)
3h
4
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TECHGURU CLASSES for SSC-JE/RAILWAYS/ORDINANCE
CHAPTER - 5 : CENTROIDS
5.3 Centroid of the arc of circle shown in figure is :
y
(a) r sin ( /2) 2  , 0
r
(b) 2r sin ( )/  , 0
x
(c) 2r sin ( /2) /  , 0
(d) r sin (  )/  , 0
Figure: 5.3
5.4 The centre of gravity of a solid cone lies on the axis at the height (h = total
height of cone) from the vertex :
(a)
h
8
(b)
3h
4
(c)
2
h
3
(d)
h
2
5.5 The centre of gravity of a hemisphere from its base measured along the
vertical (radius r) is at a distance (from the base) :
(a)
3R
8
(b)
3R
4
(c)
3
4R
(d)
4
3R
5.6 The centroid of the area between the circle x2 + y2 = 16 and the line x + y =
will have the coordinates :
(a) (4, 2)
(b)
(2, 4)
(c) (2.34, 2.34)
(d)
(1.16, 1.16)
5.7 The centroid of section of circle shown in figure is :
y
(a) 2r sin  / 3 
(b) 2r sin (  /2) 
r
(c) 2r sin (  /2) 3 
x
(d) r sin  /3 
Figure: 5.7
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TECHGURU CLASSES for SSC-JE / RAILWAYS / ORDINANCE
CHAPTER - 5 : CENTROIDS
5.8 A sphere of radius r is cut from larger sphere of radius R. The distance
between their centres is a. The centroid of the remaining volume lies on the
line of centres and the distance from the centre of the larger sphere is :
(a) aR / (R – r)
(b)
aR / (R2 – r2)
(c) ar3 / (R3 – r3)
(d)
ar2 / R2 – r2
5.9 The centre of gravity of a trapezium of base 'b', height 'h' and upper side 'a'
lies at following distance from the base :
(II)
(a)
h  2a  b 


3  ab 
(b)
h  ab 


3  2a  b 
(c)
h  a  2b 


3  ab 
(d)
h  2a  b 


3  ab 
Matching Type Questions
5.10
Column–I
Column–II
(i) Semicircular arc
(a)
h/3
(ii) Semicircular segment
(b)
2r/ 
(c)
 ab
(iii) Triangular area (right angle)
(iv) Area of elliptical quadrant
5.11
(d)
Column–I
4r / 3 
Column–II
(i) hemisphere
(a)
4r / 3 
(ii) Right circular cone
(b)
3r/8
(c)
h/4
(d)
h/4
(iii) half cone volume
(iv) Semi-circular cylinder
5.12
Column–I
Column–II
(i) Centre of gravity
(a)
Function of geometry of object.
(ii) Centroid
(b)
Function of mass distribution.
(iii) Centre of mass
(c)
Point of application of resultant
weight.
(iv) Hemogeneous solid
(d)
Centre of gravity, centre of mass
and centroid become identical.
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TECHGURU CLASSES for SSC-JE/RAILWAYS/ORDINANCE
CHAPTER - 5 : CENTROIDS
(III)
Fill in the Blanks Type Questions
5.13 The centre of gravity of trapezium of base 'b', height 'h' and upper side 'a'
lies ....................... distance from it's base.
5.14 Centre of gravity of a hemisphere lies ............................. distance from it's
centre.
5.15 Centre of gravity of right circular cone lies at .......................... distance
from it's base.
5.16 Centre of gravity of semicircle lies at ........................ distance from it's
base.
5.17 Centre of gravity of a uniform rod lies at it's .........................
5.18 If a body has a plane of symmetry centre of gravity lies ......................
plane.
(IV)
True/False Type Questions
5.19 Centre of gravity of plane lamina lies outside the body.
5.20 Centroid of a triangle lamina lies at 2h/3 distance from the side of triangle.
5.21 Centroid of a circular arc is sin  /3  .
5.22 Centroid and centre of gravity are the same point.
(V)
Theoretical Questions
5.23 Explain in brief:
(i)Centre of gravity
(ii) Centre of mass
(iii) Centroid. Also give a comparison among them.
5.24 Explain, how can you determine the centroid of a composite curve ?
5.25 Prove that centroid of a circular arc lies at a distance of
r sin 
from the

centre measured along symmetrical axis of the arc. 'r' and '2  ' are radius
of the circular arc and angle subtended by circular arc at the centre.
Hence, Determine the position of centroid of a semicircular arc.
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TECHGURU CLASSES for SSC-JE / RAILWAYS / ORDINANCE
CHAPTER - 5 : CENTROIDS
5.26 Prove that centroid of a triangle lies at
h
from the base measured along
3
the line perpendicular to base, h being the height of triangle.
5.27 Derive expressions for :
(i)Centroid of homogeneous semicylinder from its base measured along
an axis perpendicular to the base.
(ii) Centroid of a homogeneous hemisphere from its base measured along
an axis perpendicular to the base.
(iii) Centroid of a homogeneous right circular cone from its base measured
along an axis perpendicular to the base.
(VI)
Numerical Questions
Determine centroids of composite figures shown in figures for questions
5.28 to 5.32.
5.28
y
10
All dimensions are in 'cm'.
25
x
20
Figure: 5.28
5.29
10
25
All dimensions are in 'mm'.
20
x
50
Figure: 5.29
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TECHGURU CLASSES for SSC-JE/RAILWAYS/ORDINANCE
CHAPTER - 5 : CENTROIDS
y
5.30
A
50mm
C
O
D
x
B
Given: OD = 65 mm.
y
5.31
C
ABCD is a square of side 'a'.
D
B
x
A
y
5.32
B
C
5
30
O
A
50
x
ANSWERS
5.1 (b)
5.2
(c)
5.3
(d)
5.4
(b)
5.6 (c)
5.7
(a)
5.8
(c)
5.9
(a)
5.5
(a)
5.10 [(i) – (b), (ii) – (d), (iii) – (a), (iv) – (c)]
5.11 [(i) – (d), (ii) – (c), (iii) – (d), (iv) – (a)]
5.12 [(i) – (c), (ii) – (a), (iii) – (b), (iv) – (d)]
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TECHGURU CLASSES for SSC-JE / RAILWAYS / ORDINANCE
CHAPTER - 5 : CENTROIDS
h (2a  b)
5.13 3 (a  b)
5.14
3r
8
5.15
h
4
5.16
4r
3
5.17
Mid point
5.18
on same.
5.19
False
5.20
False
5.21
False
5.22
False
5.23
a
5.24
a
5.25
a
5.26
a
5.27
a
5.28
x = 0,
5.30
x = 47.81, y = 6.93
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y = 16.28 cm
5.29
y = 31.79 mm
5.31
x  y = 29.63 units
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